Emma's Dilemma

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Introduction

Maths Coursework

Emma’s Dilemma

I will be investigating the different arrangements of letters of words, which don’t have any identical letters in them and those that do. Then I will try and find a formula that can calculate the total arrangements of letters of any word, which does not have any identical letters in it. I will also try and find a formula to find the total arrangements of words, which have some identical letters in them.

Firstly I will look at those words, which have no identical letters in them and then work out a formula that can find the total arrangements of letters in them. So I will begin by finding the total arrangements of a one-lettered word to finding the total arrangements of a four-lettered word. Thereafter I will try and predict the total arrangements of a five-lettered word and then I will check if I was correct by finding the total arrangements of it’s letters

Part 1

One-lettered word

A =1

There is only one arrangement for a one-lettered word.

Two-lettered word

WE

EW =2

There are two different arrangements for a two-lettered word whose letters are all different.

Three-lettered word

CAR CRA

ACR ARC =6

RCA RAC

There are six different arrangement for a three-lettered word whose letters are all different.

Four-lettered word

LUCY LUYC LYUC LYCU LCYU LCUY

ULCY ULYC UCLY UCYL UYCL UYLC

CLUY CLYU CULY CUYL CYUL CYLU

YUCL YULC YCUL YCLU YLCU YLUC =24

There are 24 different arrangement for a four-lettered word whose letters are all different.

It is apparent that as the number of letters in a word increase the number of different arrangements increase. When there is a two-lettered word the total arrangement is 2, which is worked out by doing 2 x 1 and 1 is also the total number of arrangement for a one-lettered. So I think that to find the total arrangements of a word you have to multiply the total number of arrangements of the word before it by the number of letters in that word. This theory can be confirmed by multiplying 24 (the total number of arrangements of a four-lettered word) by 5 (the number of letters in that word) which equals 120. But to find the total arrangements of a four lettered-word you have to multiply 1, which is the total arrangement of a one-lettered word by 2 and then multiply that by 3, which gives you 6. Then you multiply 6 by 4 to find the total arrangements of a four-lettered word. You can put all this together like this: 1x2x3x4x5 = 120.

So to find the total arrangements of any letter you have to start by multiplying from 1 to the number of letters of that word. So to find the total arrangements of a seven-lettered word you do this: 1x2x3x4x5x6x7, which equals 5040. This is called factorial.

Factorial is known by ! which can be found on a scientific calculator.

So the formula to find the total arrangement of a word, which has no repeated letters in it is n!.

Part 2

I will now investigate the total arrangements of words, which have some repeated letters in them. Then I will try and find a formula for finding the total arrangement.

First I will investigate the total arrangements of words, which have a letter repeated in it twice.

Two-lettered word with a letter repeated twice

AA = 1

There is only one arrangement for a two-lettered word, which has a letter repeated in it twice.

Three-lettered word with a letter repeated twice

RAA ARA AAR = 3

There are only three different arrangements for a three-lettered word, which has a letter repeated in it twice.

Four-lettered word with a letter repeated twice

EMMA EAMM EMAM

MEAM MAEM MAME

MMEA MMAE MEMA

AEMM AMEM AMME = 12

There are 12 different arrangements for a four-lettered word, which a letter repeated in it twice.

Five-lettered word with a letter repeated twice

SLEEP SLEPE SLPEE SPLEE SPELE SPEEL

SEEPL SEPLE SELPE SEPEL SEELP SELEP

LSEEP LSEPE LSPEE LPEES LPESE LPSEE

LEEPS LEESP LESPE LESEP LEPES LEPSE

PSLEE PSELE PSEEL PLSEE PLESE PLEES

PEELS PEESL PESEL PESLE PELSE PELES

ESPEL ESPLE ESEPL ESELP ESLEP ESLPE

EPSEL EPSLE EPLSE EPLES EPELS EPESL

ELSPE ELSEP ELPSE ELPES ELEPS ELESP

EELSP EELPS EEPLS EEPSL EESPL EESLP = 60

There are 60 different arrangements for a five-lettered word, which has a letter repeated

Here is a table showing the total arrangements of words, which have a letter repeated in it twice.

Number of Letters

Total arrangement

2

1

3

3

4

12

5

60

If you compare this with the total arrangements of words, which have no letters repeated in them you will notice that the total arrangement of a word, which has a letter repeated twice is half of the arrangement of a word, which has the same number of letters but has no letter repeated twice.

So to find the formula I would have to use the formula for a word, which has no letter repeated which is n! and then divide it by 2. So when you divide 24 (which is the total arrangement of LUCY) you get 12 (which is the total arrangement of EMMA).

So the formula to find the total arrangement of a word, which has a letter repeated in it twice is (n!)/2.

Now I will investigate the total arrangements of words, which have a letter repeated three times.

Three-lettered word which has a letter repeated three times

TTT = 1

There is only one arrangement for a three-lettered word, which has a letter repeated three times.

Four-lettered word which has a letter repeated three times

STTT TSTT TTST TTTS = 4

There are four different arrangements for a four-lettered word, which has a letter repeated in it three times.

Five-lettered word which has a letter repeated three times

ASTTT ATSTT ATTST ATTTS

SATTT STATT STTAT STTTA

TASTT TATST TATTS TSATT

TSTAT TSTTA TTSAT TTSTA

TTAST TTATS TTTAS TTTSA = 20

There are 2o different arrangements for a five-lettered word, which has a letter repeated three times.

Here is a table showing the total arrangements of words, which have a letter repeated three times.

Four-lettered word which has two types of letters each repeated twice

There are 6 different arrangements of a four-lettered word, which has two different types of letters each repeated twice.

The formula for a four-lettered word, which has only two types of letters is n!/4.

You get 4 when you multiply 2! by 2! So I think that the formula to find the total arrangement of a word, which has two types of letters repeated two times or more is (n!)/(r!)(x!) (x stands for the other letter which is also repeated)

So to find the total arrangement of a four-lettered word, which has two types of letters repeated twice, you would do this: 4!

(2!)(2!)

= 24

4

= 6

But if there are more than two types of letters repeated all you do is add more factorials according the number of types of different letters that are repeated.

So to find the total arrangement of RRHHAA you will do this: 6!

(2!)(2!)(2!)

You have three factorials because there are three different types of letters repeated twice

= 720

8

= 90

So the total arrangement of RRHHAA is 90.

So the formula to find the total arrangement of a word, which has more than one type of letter repeated is (n!)/(Factorials) (the number of factorials depend on the number of different types of letters repeated)

Here are all the formulas that I have found from this investigation.

n! When there are no letters repeated.

(n!)/(r!) When there is a letter repeated 2 or more times.

(n)/(Factorials) When there are more than one type of letter repeated.

From this investigation I have found out many formulas, which will help me to find the total arrangements of different types of words. Finding the formulas weren’t too difficult but trying to put them into words was a bit hard.

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This means that the total number of arrangements possible is reduced by a factor of 4, or "2 ! X 2!" ( See later ). For example For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.

212211 total arrangement is 20 let's see the formula: a=(1*2*3*4*5*6)/1*2*3 =120 no, it doesn't work but if we divide it by another 6 which is (1*2*3) Can you see the pattern, the formula still work if we times the mutiply again.

Let's confirm this formula, if a nuber has 1fig it has 1 arrangements formular: 1*1=1 It works 2fig it has 2 arrangements formular: 1*2=2 It works 3fig it has 6 arrangements formular: 1*2*3=6 It works 4fig it has 24 arrangements formular: 1*2*3*4=24 It works Formula is confirmed What about if

of letters - 1 ) X ( No. of letters - 2 ) X etc... combinations No. of times the letter has been repeated For example: To find the number of combinations which can be made from using 10 letters, with two of the ten being the same, can be found by using the following formula: No.

formular is confirmed This formula can be written as: a=ni/xixi x represent the number of figures of same number What about a number with difference number of figure of same number. For example: 11122,111122 let's try if the formula still work.

the numbers of arrangements for 2 and 3 have been missed out. Looking at the table the original combination has been divided by 6 and 2 x 3 = 6. so in a matter of fact the number of combinations is n!/3x2, 3x2 is really the same as 3!

First, I will investigate a 3-letter word. CCC There is only 1 arrangement. Now I am going to investigate a 4-letter word. AAAR AARA ARAA RAAA There are 4 arrangements. Now I will investigate a 5-letter word. YYYEL YYYLE YYEYL YYELY YYLEY YYLYE YEYYL YEYLY YELYY YLYYE YLYEY YLEYY EYYYL